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Prove whethere is linearly independent or not the following:
{[tex]e^x,e^{2x}[/tex]}
{[tex]e^x,e^{2x}[/tex]}
I think it's possible:
[tex]a e^x + b e^{2x} = 0[/tex] <==> [tex]a e^x = - b e^{2x}[/tex] <==> [tex]ln (a e^x) = - ln (b e^{2x})[/tex] <==> ax = -2bx
So above, if we carefully choose a, we can definitely show your given set is linearly dependent.
Except
[tex]\ln(a e^x) = \ln(a) + x[/tex]
[tex]\ln(-b e^{2x}) = \ln(-b) + 2x[/tex]
so you need
[tex]\ln(a) - \ln(-b) = x[/tex] for all x!
If not, then there must exist two scalars, a and b (they can't both be zero), such that
[tex]a e^x + b e^{2x} = 0[/tex]
for all x. Can you find such scalars?
No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.
No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.Put x=0,then a+b=0 ====> a=-b
Hence the above are linearly dependent
CORRECT???
THANKS
No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.
Another way to prove it is this: if [itex]ae^x+ be^{2x}= 0[/itex] for all x, then it is a constant and its derivative must be 0 for all x: [itex]ae^x+ 2be^{2x}= 0[/itex]. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.
Is not the definition for linear indepedence the following:
for all a,b and x ,real Nos and [tex]ae^x + be^{2x}=0\Longrightarrow a=0=b[/tex]
Yes, that is why you can't "prove [itex]e^x[/itex] and [itex]e^{2x}[/itex] are dependent": they are independent.
That isn't right. You have your quantifiers all kludged up into one it should have read
(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)
.
That isn't right. You have your quantifiers all kludged up into one it should have read
(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)
the negation of which is
(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))
The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.
You did not quote all of what he said. He said "The for all x is not part of the definition of linear (in)dependence" and then followed with "it is part of the definition of what it means for the function to be zero" and that was the point of my last response:The negation of :
(for all a,b)(ae^x + be^2x =0for all x ===.> a=b=0) is
(there exist a,b) [~(ae^x + be^2x = 0 for all x ====> a=b=0)] ======>
(there exist a,b) [ ae^x = be^2x =0 there exist x AND a=/=0 ,b=/=0]
Whether you put the quantifier infront ,like i did, or at the end ,like you did we still have the same result.
The negation of [tex]\forall xPx[/tex] is : [tex]\exists x\neg Px[/tex] .
In our case Px is ( ae^x + be^2x =0 ======> a=b=0) and the negation of Px is:
ae^x = be^2x =0 and a=/=0,b=/=0
When you want to show that the function ae^x +be^2 is zero for all xεR ,YOU write :
for all xεR, ae^x + be^2x =0 OR ae^x + be^2x = 0, for all x.
To say that: for all x, is not part of the definition ,then what is part of??
You did not quote all of what he said. He said "The for all x is not part of the definition of linear (in)dependence" and then followed with "it is part of the definition of what it means for the function to be zero" and that was the point of my last response:
We are talking about functions of x. Saying that "[itex]ae^x+ be^{-x}= 0[/itex]" means that the function [itex]f(x)= ae^x+ be^{-x}[/itex] is equal to the "0 function": g(x)= 0.
: can you find real numbers a and b not both equal to 0 so that
ae^x + be^2x
is the zero function? No - it has been proven several times in this thread.
How do you get that last step? Did you set [itex]a= -e^x[/itex] and b= 1? Of course, you can't do that. [itex]-e^x[/itex] is not a constant.O.K here is a proof having the for all x part:
Let b=1
Let a= -e^x
AND [tex]ae^x + be^{2x} = e^x( a + be^x) = e^x( -e^x + e^x)[/tex]
[tex] = e^x.0 = 0[/tex] [tex]for all x\in R[/tex] >
Hence we have proved there exist [tex]a\neq 0,b\neq 0[/tex] and such that :
[tex] ae^x + be^{2x} = 0[/tex] for all x
In what way is that not in a useful quantifier form (for those who like things like that)? Really, we shouldn't even bother with the quantifier "for all a,b", and putting things in the full on formal abstract quantifier notation just makes things far more opaque than they need to be.
You're making a very easy question seem very hard: can you find real numbers a and b not both equal to 0 so that
ae^x + be^2x
is the zero function? No - it has been proven several times in this thread.
I can only conclude that you have not understood anything anyone has said here. matt grime is saying clearly here that it has been proven several time in this thread that it is impossible to find such a and b. Your statement here is not at all responsive to that.If you cannot find them it does not mean they do not exist
I can only conclude that you have not understood anything anyone has said here. matt grime is saying clearly here that it has been proven several time in this thread that it is impossible to find such a and b. Your statement here is not at all responsive to that.